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Using Feynman's technique, I want to evaluate $$\int_0^{\infty}\frac{\cos{x}}{x^2+1} dx$$

I set

$$I(a)=\int_0^{\infty}\frac{\cos{ax}}{x^2+1} dx$$

which gives

$$I'(a)=\int_0^{\infty}\frac{-x\sin{ax}}{x^2+1} dx$$

$$I''(a)=\int_0^{\infty}\frac{-x^2\cos{ax}}{x^2+1} dx$$

However, when I try to form a differential equation, I get

$$I(a)-I''(a)=\int_0^{\infty}\cos{ax}\; dx$$

However, this integral doesn't converge and I'm stuck on how I can work my way around this. I know that to get the right answer, I need $I(a)-I''(a)=0$ but I can't get there. Can someone help me find the right way to do this question?

Rodrigo de Azevedo
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Pen and Paper
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    The second derivative does not converge as an integral, as you have noted, so you may not use it. Apply IBP on the first derivative maybe – FShrike May 17 '22 at 11:44

1 Answers1

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The trick for the problem is to take the first derivative integral and multiply the integrand by $\frac{x}{x}$:

$$\int_{0}^{\infty} \frac{-x^2}{x^2+1}\cdot\frac{\sin ax}{x}\:dx = -\int_0^\infty\frac{\sin ax}{x}\:dx+ \int_0^\infty \frac{\sin ax}{x(x^2+1)}\:dx$$

$$ \implies I'(a) = -\frac{\pi}{2}+ \int_0^\infty \frac{\sin ax}{x(x^2+1)}\:dx$$

Then we get

$$I''(a) = \int_0^\infty \frac{\cos ax}{x^2+1}\:dx = I(a)$$

Ninad Munshi
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